Proposition: What the present calculation does not decide

The proved results establish neither of the decisive E1 conclusions:

1. a zero joint-symbol section for the complete six-branch coefficient with controlled remainder;

2. a nonzero joint-symbol section of that complete coefficient.

They also do not demonstrate an incompatibility excluding every prescription under the frozen contract.

Proof. Theorem thm:regroup is an exact identity for the already defined smooth-slot tree kernels. It does not compose them with the cross-lines on the characteristic set.

Theorem thm:nonclean rules out the elementary phase candidate, not all possible representations. Proposition prop:quadratic identifies its geometric degeneracy but supplies no distributional amplitude estimate.

The copied $Q\lambda$ identity concerns only $B_6$. None of these statements supplies a common joint-symbol section or a remainder theorem for (eq:target). Nor does any statement quantify over all prescriptions in a previously fixed candidate class.